Sponsored Links

Jumat, 03 November 2017

Sponsored Links

Separation of Variables - Heat Equation Part 1 - YouTube
src: i.ytimg.com

The heat equation is a parabolic partial differential equation that describes the distribution of heat (or variation in temperature) in a given region over time.


Video Heat equation



Statement of the equation

For a function u(x,y,z,t) of three spatial variables (x,y,z) (see cartesian coordinates) and the time variable t, the heat equation is

? u ? t - ? ( ? 2 u ? x 2 + ? 2 u ? y 2 + ? 2 u ? z 2 ) = 0 {\displaystyle {\frac {\partial u}{\partial t}}-\alpha \left({\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}+{\frac {\partial ^{2}u}{\partial z^{2}}}\right)=0}

More generally in any coordinate system:

where ? is a positive constant, and ? or ?2 denotes the Laplace operator. In the physical problem of temperature variation, u(x,y,z,t) is the temperature and ? is the thermal diffusivity. For the mathematical treatment it is sufficient to consider the case ? = 1.

Note that the state equation, given by the first law of thermodynamics (i.e. conservation of energy), is written in the following form (assuming no mass transfer or radiation). This form is more general and particularly useful to recognize which property (e.g. cp or ? {\displaystyle \rho } ) influences which term.

? c p ? T ? t - ? ? ( k ? T ) = q ? V {\displaystyle \rho c_{p}{\frac {\partial T}{\partial t}}-\nabla \cdot \left(k\nabla T\right)={\dot {q}}_{V}}

where q ? V {\displaystyle {\dot {q}}_{V}} is the volumetric heat source.

The heat equation is of fundamental importance in diverse scientific fields. In mathematics, it is the prototypical parabolic partial differential equation. In probability theory, the heat equation is connected with the study of Brownian motion via the Fokker-Planck equation. In financial mathematics it is used to solve the Black-Scholes partial differential equation. The diffusion equation, a more general version of the heat equation, arises in connection with the study of chemical diffusion and other related processes.


Maps Heat equation



General description

Suppose one has a function u that describes the temperature at a given location (x, y, z). This function will change over time as heat spreads throughout space. The heat equation is used to determine the change in the function u over time. The rate of change of u is proportional to the "curvature" of u. Thus, the sharper the corner, the faster it is rounded off. Over time, the tendency is for peaks to be eroded, and valleys filled in. If u is linear in space (or has a constant gradient) at a given point, then u has reached steady-state and is unchanging at this point (assuming a constant thermal conductivity).

The image to the right is animated and describes the way heat changes in time along a metal bar. One of the interesting properties of the heat equation is the maximum principle that says that the maximum value of u is either earlier in time than the region of concern or on the edge of the region of concern. This is essentially saying that temperature comes either from some source or from earlier in time because heat permeates but is not created from nothingness. This is a property of parabolic partial differential equations and is not difficult to prove mathematically (see below).

Another interesting property is that even if u has a discontinuity at an initial time t = t0, the temperature becomes smooth as soon as t > t0. For example, if a bar of metal has temperature 0 and another has temperature 100 and they are stuck together end to end, then very quickly the temperature at the point of connection will become 50 and the graph of the temperature will run smoothly from 0 to 50.

The heat equation is used in probability and describes random walks. It is also applied in financial mathematics for this reason.

It is also important in Riemannian geometry and thus topology: it was adapted by Richard S. Hamilton when he defined the Ricci flow that was later used by Grigori Perelman to solve the topological Poincaré conjecture.


pde - Heat equation, boundary gradient singularity - Mathematics ...
src: i.stack.imgur.com


The physical problem and the equation

Derivation in one dimension

The heat equation is derived from Fourier's law and conservation of energy (Cannon 1984). By Fourier's law, the rate of flow of heat energy per unit area through a surface is proportional to the negative temperature gradient across the surface,

q = - k ? u   {\displaystyle \mathbf {q} =-k\,\nabla u\ }

where k is the thermal conductivity and u is the temperature. In one dimension, the gradient is an ordinary spatial derivative, and so Fourier's law is

q = - k ? u ? x {\displaystyle q=-k{\frac {\partial u}{\partial x}}\,}

In the absence of work done, a change in internal energy per unit volume in the material, ?Q, is proportional to the change in temperature, ?u (in this section only, ? is the ordinary difference operator with respect to time, not the Laplacian with respect to space). That is,

? Q = c p ? ? u {\displaystyle \Delta Q=c_{p}\rho \,\Delta u\,}

where cp is the specific heat capacity and ? is the mass density of the material. Choosing zero energy at absolute zero temperature, this can be rewritten as

Q = c p ? u . {\displaystyle Q=c_{p}\rho u.\,}

The increase in internal energy in a small spatial region of the material

x - ? x <= ? <= x + ? x {\displaystyle x-\Delta x\leq \xi \leq x+\Delta x}

over the time period

t - ? t <= ? <= t + ? t {\displaystyle t-\Delta t\leq \tau \leq t+\Delta t}

is given by

c p ? ? x - ? x x + ? x [ u ( ? , t + ? t ) - u ( ? , t - ? t ) ] d ? = c p ? ? t - ? t t + ? t ? x - ? x x + ? x ? u ? ? d ? d ? {\displaystyle c_{p}\rho \int _{x-\Delta x}^{x+\Delta x}[u(\xi ,t+\Delta t)-u(\xi ,t-\Delta t)]\,d\xi =c_{p}\rho \int _{t-\Delta t}^{t+\Delta t}\int _{x-\Delta x}^{x+\Delta x}{\frac {\partial u}{\partial \tau }}\,d\xi \,d\tau }

where the fundamental theorem of calculus was used. If no work is done and there are neither heat sources nor sinks, the change in internal energy in the interval [x-?x, x+?x] is accounted for entirely by the flux of heat across the boundaries. By Fourier's law, this is

k ? t - ? t t + ? t [ ? u ? x ( x + ? x , ? ) - ? u ? x ( x - ? x , ? ) ] d ? = k ? t - ? t t + ? t ? x - ? x x + ? x ? 2 u ? ? 2 d ? d ? {\displaystyle k\int _{t-\Delta t}^{t+\Delta t}\left[{\frac {\partial u}{\partial x}}(x+\Delta x,\tau )-{\frac {\partial u}{\partial x}}(x-\Delta x,\tau )\right]\,d\tau =k\int _{t-\Delta t}^{t+\Delta t}\int _{x-\Delta x}^{x+\Delta x}{\frac {\partial ^{2}u}{\partial \xi ^{2}}}\,d\xi \,d\tau }

again by the fundamental theorem of calculus. By conservation of energy,

? t - ? t t + ? t ? x - ? x x + ? x [ c p ? u ? - k u ? ? ] d ? d ? = 0. {\displaystyle \int _{t-\Delta t}^{t+\Delta t}\int _{x-\Delta x}^{x+\Delta x}[c_{p}\rho u_{\tau }-ku_{\xi \xi }]\,d\xi \,d\tau =0.}

This is true for any rectangle [t -?t, t + ?t] × [x - ?x, x + ?x]. By the fundamental lemma of the calculus of variations, the integrand must vanish identically:

c p ? u t - k u x x = 0. {\displaystyle c_{p}\rho u_{t}-ku_{xx}=0.\,\!}

Which can be rewritten as:

u t = k c p ? u x x , {\displaystyle u_{t}={\frac {k}{c_{p}\rho }}u_{xx},}

or:

? u ? t = k c p ? ( ? 2 u ? x 2 ) {\displaystyle {\frac {\partial u}{\partial t}}={\frac {k}{c_{p}\rho }}\left({\frac {\partial ^{2}u}{\partial x^{2}}}\right)}

which is the heat equation, where the coefficient (often denoted ?)

? = k c p ? {\displaystyle \alpha ={\frac {k}{c_{p}\rho }}\,\!}

is called the thermal diffusivity.

An additional term may be introduced into the equation to account for radiative loss of heat, which depends upon the excess temperature u = T - Ts at a given point compared with the surroundings. At low excess temperatures, the radiative loss is approximately ?u, giving a one-dimensional heat-transfer equation of the form

? u ? t = k c p ? ( ? 2 u ? x 2 ) - ? u . {\displaystyle {\frac {\partial u}{\partial t}}={\frac {k}{c_{p}\rho }}\left({\frac {\partial ^{2}u}{\partial x^{2}}}\right)-\mu u.}

At high excess temperatures, however, the Stefan-Boltzmann law gives a net radiative heat-loss proportional to T 4 - T s 4 {\displaystyle T^{4}-T_{s}^{4}} , and the above equation is inaccurate. For large excess temperatures, T 4 - T s 4 ? u 4 {\displaystyle T^{4}-T_{s}^{4}\approx u^{4}} , giving a high-temperature heat-transfer equation of the form

? u ? t = ? ( ? 2 u ? x 2 ) -   m u 4 {\displaystyle {\frac {\partial u}{\partial t}}=\alpha \left({\frac {\partial ^{2}u}{\partial x^{2}}}\right)-\ mu^{4}}

where   m = ? ? p / ? A c p {\displaystyle \ m=\epsilon \sigma p/\rho Ac_{p}} . Here, ? is Stefan's constant, ? is a characteristic constant of the material, p is the sectional perimeter of the bar and A is its cross-sectional area. However, using T instead of u gives a better approximation in this case.

Three-dimensional problem

In the special cases of wave propagation of heat in an isotropic and homogeneous medium in a 3-dimensional space, this equation is

? u ? t = ? ? 2 u = ? ( ? 2 u ? x 2 + ? 2 u ? y 2 + ? 2 u ? z 2 ) {\displaystyle {\partial u \over \partial t}=\alpha \nabla ^{2}u=\alpha \left({\partial ^{2}u \over \partial x^{2}}+{\partial ^{2}u \over \partial y^{2}}+{\partial ^{2}u \over \partial z^{2}}\right)} = ? ( u x x + u y y + u z z ) {\displaystyle =\alpha (u_{xx}+u_{yy}+u_{zz})\quad }

where:

  • u = u(x, y, z, t) is temperature as a function of space and time;
  • ? u ? t {\displaystyle {\frac {\partial u}{\partial t}}} is the rate of change of temperature at a point over time;
  • uxx, uyy, and uzz are the second spatial derivatives (thermal conductions) of temperature in the x, y, and z directions, respectively;
  • ? = k c p ? {\displaystyle \alpha ={\frac {k}{c_{p}\rho }}} is the thermal diffusivity, a material-specific quantity depending on the thermal conductivity k, the mass density ?, and the specific heat capacity cp.

The heat equation is a consequence of Fourier's law of conduction (see heat conduction).

If the medium is not the whole space, in order to solve the heat equation uniquely we also need to specify boundary conditions for u. To determine uniqueness of solutions in the whole space it is necessary to assume an exponential bound on the growth of solutions.

Solutions of the heat equation are characterized by a gradual smoothing of the initial temperature distribution by the flow of heat from warmer to colder areas of an object. Generally, many different states and starting conditions will tend toward the same stable equilibrium. As a consequence, to reverse the solution and conclude something about earlier times or initial conditions from the present heat distribution is very inaccurate except over the shortest of time periods.

The heat equation is the prototypical example of a parabolic partial differential equation.

Using the Laplace operator, the heat equation can be simplified, and generalized to similar equations over spaces of arbitrary number of dimensions, as

u t = ? ? 2 u = ? ? u , {\displaystyle u_{t}=\alpha \nabla ^{2}u=\alpha \Delta u,\quad \,\!}

where the Laplace operator, ? or ?2, the divergence of the gradient, is taken in the spatial variables.

The heat equation governs heat diffusion, as well as other diffusive processes, such as particle diffusion or the propagation of action potential in nerve cells. Although they are not diffusive in nature, some quantum mechanics problems are also governed by a mathematical analog of the heat equation (see below). It also can be used to model some phenomena arising in finance, like the Black-Scholes or the Ornstein-Uhlenbeck processes. The equation, and various non-linear analogues, has also been used in image analysis.

The heat equation is, technically, in violation of special relativity, because its solutions involve instantaneous propagation of a disturbance. The part of the disturbance outside the forward light cone can usually be safely neglected, but if it is necessary to develop a reasonable speed for the transmission of heat, a hyperbolic problem should be considered instead - like a partial differential equation involving a second-order time derivative. Some models of nonlinear heat conduction (which are also parabolic equations) have solutions with finite heat transmission speed.

Internal heat generation

The function u above represents temperature of a body. Alternatively, it is sometimes convenient to change units and represent u as the heat density of a medium. Since heat density is proportional to temperature in a homogeneous medium, the heat equation is still obeyed in the new units.

Suppose that a body obeys the heat equation and, in addition, generates its own heat per unit volume (e.g., in watts/litre - W/L) at a rate given by a known function q varying in space and time. Then the heat per unit volume u satisfies an equation

? u ? t = ? ( ? 2 u ? x 2 + ? 2 u ? y 2 + ? 2 u ? z 2 ) + 1 c p ? q . {\displaystyle {\frac {\partial u}{\partial t}}=\alpha \left({\partial ^{2}u \over \partial x^{2}}+{\partial ^{2}u \over \partial y^{2}}+{\partial ^{2}u \over \partial z^{2}}\right)+{\frac {1}{c_{p}\rho }}q.}

For example, a tungsten light bulb filament generates heat, so it would have a positive nonzero value for q when turned on. While the light is turned off, the value of q for the tungsten filament would be zero.


2D Heat Equation (MATLAB) - YouTube
src: i.ytimg.com


Solving the heat equation using Fourier series

The following solution technique for the heat equation was proposed by Joseph Fourier in his treatise Théorie analytique de la chaleur, published in 1822. Let us consider the heat equation for one space variable. This could be used to model heat conduction in a rod. The equation is

where u = u(x, t) is a function of two variables x and t. Here

  • x is the space variable, so x ? [0, L], where L is the length of the rod.
  • t is the time variable, so t >= 0.

We assume the initial condition

where the function f is given, and the boundary conditions

Let us attempt to find a solution of (1) that is not identically zero satisfying the boundary conditions (3) but with the following property: u is a product in which the dependence of u on x, t is separated, that is:

This solution technique is called separation of variables. Substituting u back into equation (1),

T ? ( t ) ? T ( t ) = X ? ( x ) X ( x ) . {\displaystyle {\frac {T'(t)}{\alpha T(t)}}={\frac {X''(x)}{X(x)}}.}

Since the right hand side depends only on x and the left hand side only on t, both sides are equal to some constant value -?. Thus:

and

We will now show that nontrivial solutions for (6) for values of ? <= 0 cannot occur:

This solves the heat equation in the special case that the dependence of u has the special form (4).

In general, the sum of solutions to (1) that satisfy the boundary conditions (3) also satisfies (1) and (3). We can show that the solution to (1), (2) and (3) is given by

u ( x , t ) = ? n = 1 ? D n sin ( n ? x L ) e - n 2 ? 2 ? t L 2 {\displaystyle u(x,t)=\sum _{n=1}^{\infty }D_{n}\sin \left({\frac {n\pi x}{L}}\right)e^{-{\frac {n^{2}\pi ^{2}\alpha t}{L^{2}}}}}

where

D n = 2 L ? 0 L f ( x ) sin ( n ? x L ) d x . {\displaystyle D_{n}={\frac {2}{L}}\int _{0}^{L}f(x)\sin \left({\frac {n\pi x}{L}}\right)\,dx.}

Generalizing the solution technique

The solution technique used above can be greatly extended to many other types of equations. The idea is that the operator uxx with the zero boundary conditions can be represented in terms of its eigenvectors. This leads naturally to one of the basic ideas of the spectral theory of linear self-adjoint operators.

Consider the linear operator ?u = uxx. The infinite sequence of functions

e n ( x ) = 2 L sin ( n ? x L ) {\displaystyle e_{n}(x)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi x}{L}}\right)}

for n >= 1 are eigenvectors of ?. Indeed,

? e n = - n 2 ? 2 L 2 e n . {\displaystyle \Delta e_{n}=-{\frac {n^{2}\pi ^{2}}{L^{2}}}e_{n}.}

Moreover, any eigenvector f of ? with the boundary conditions f(0) = f(L) = 0 is of the form en for some n >= 1. The functions en for n >= 1 form an orthonormal sequence with respect to a certain inner product on the space of real-valued functions on [0, L]. This means

? e n , e m ? = ? 0 L e n ( x ) e m * ( x ) d x = ? m n {\displaystyle \langle e_{n},e_{m}\rangle =\int _{0}^{L}e_{n}(x)e_{m}^{*}(x)dx=\delta _{mn}}

Finally, the sequence {en}n ? N spans a dense linear subspace of L2((0, L)). This shows that in effect we have diagonalized the operator ?.


Heat Equation â€
src: hnagibdotcom1.files.wordpress.com


Heat conduction in non-homogeneous anisotropic media

In general, the study of heat conduction is based on several principles. Heat flow is a form of energy flow, and as such it is meaningful to speak of the time rate of flow of heat into a region of space.

  • The time rate of heat flow into a region V is given by a time-dependent quantity qt(V). We assume q has a density Q {\displaystyle Q} , so that
q t ( V ) = ? V Q ( x , t ) d x {\displaystyle q_{t}(V)=\int _{V}Q(x,t)\,dx\quad }
  • Heat flow is a time-dependent vector function H(x) characterized as follows: the time rate of heat flowing through an infinitesimal surface element with area dS and with unit normal vector n is
H ( x ) ? n ( x ) d S {\displaystyle \mathbf {H} (x)\cdot \mathbf {n} (x)\,dS}
Thus the rate of heat flow into V is also given by the surface integral
q t ( V ) = - ? ? V H ( x ) ? n ( x ) d S {\displaystyle q_{t}(V)=-\int _{\partial V}\mathbf {H} (x)\cdot \mathbf {n} (x)\,dS}
where n(x) is the outward pointing normal vector at x.
  • The Fourier law states that heat energy flow has the following linear dependence on the temperature gradient
H ( x ) = - A ( x ) ? ? u ( x ) {\displaystyle \mathbf {H} (x)=-\mathbf {A} (x)\cdot \nabla u(x)}
where A(x) is a 3 × 3 real matrix that is symmetric and positive definite.
  • By the divergence theorem, the previous surface integral for heat flow into V can be transformed into the volume integral
q t ( V ) = - ? ? V H ( x ) ? n ( x ) d S = ? ? V A ( x ) ? ? u ( x ) ? n ( x ) d S = ? V ? i , j ? x i ( a i j ( x ) ? x j u ( x , t ) ) d x {\displaystyle {\begin{aligned}q_{t}(V)&=-\int _{\partial V}\mathbf {H} (x)\cdot \mathbf {n} (x)\,dS\\&=\int _{\partial V}\mathbf {A} (x)\cdot \nabla u(x)\cdot \mathbf {n} (x)\,dS\\&=\int _{V}\sum _{i,j}\partial _{x_{i}}{\bigl (}a_{ij}(x)\partial _{x_{j}}u(x,t){\bigr )}\,dx\end{aligned}}}
  • The time rate of temperature change at x is proportional to the heat flowing into an infinitesimal volume element, where the constant of proportionality is dependent on a constant ?
? t u ( x , t ) = ? ( x ) Q ( x , t ) {\displaystyle \partial _{t}u(x,t)=\kappa (x)Q(x,t)\,}

Putting these equations together gives the general equation of heat flow:

? t u ( x , t ) = ? ( x ) ? i , j ? x i ( a i j ( x ) ? x j u ( x , t ) ) {\displaystyle \partial _{t}u(x,t)=\kappa (x)\sum _{i,j}\partial _{x_{i}}{\bigl (}a_{ij}(x)\partial _{x_{j}}u(x,t){\bigr )}}

Remarks.

  • The coefficient ?(x) is the inverse of specific heat of the substance at x × density of the substance at x: ?= 1 / ( ? c p ) {\displaystyle 1/(\rho c_{p})} .
  • In the case of an isotropic medium, the matrix A is a scalar matrix equal to thermal conductivity k {\displaystyle k} .
  • In the anisotropic case where the coefficient matrix A is not scalar and/or if it depends on x, then an explicit formula for the solution of the heat equation can seldom be written down. Though, it is usually possible to consider the associated abstract Cauchy problem and show that it is a well-posed problem and/or to show some qualitative properties (like preservation of positive initial data, infinite speed of propagation, convergence toward an equilibrium, smoothing properties). This is usually done by one-parameter semigroups theory: for instance, if A is a symmetric matrix, then the elliptic operator defined by
A u ( x ) := ? i , j ? x i a i j ( x ) ? x j u ( x ) {\displaystyle Au(x):=\sum _{i,j}\partial _{x_{i}}a_{ij}(x)\partial _{x_{j}}u(x)}
is self-adjoint and dissipative, thus by the spectral theorem it generates a one-parameter semigroup.

26-Solving 1D heat equation with zero-temperature boundaries - YouTube
src: i.ytimg.com


Fundamental solutions

A fundamental solution, also called a heat kernel, is a solution of the heat equation corresponding to the initial condition of an initial point source of heat at a known position. These can be used to find a general solution of the heat equation over certain domains; see, for instance, (Evans 1998) for an introductory treatment.

In one variable, the Green's function is a solution of the initial value problem

{ u t ( x , t ) - k u x x ( x , t ) = 0 ( x , t ) ? R × ( 0 , ? ) u ( x , 0 ) = ? ( x ) {\displaystyle {\begin{cases}u_{t}(x,t)-ku_{xx}(x,t)=0&(x,t)\in \mathbf {R} \times (0,\infty )\\u(x,0)=\delta (x)&\end{cases}}}

where ? is the Dirac delta function. The solution to this problem is the fundamental solution

? ( x , t ) = 1 4 ? k t exp ( - x 2 4 k t ) . {\displaystyle \Phi (x,t)={\frac {1}{\sqrt {4\pi kt}}}\exp \left(-{\frac {x^{2}}{4kt}}\right).}

One can obtain the general solution of the one variable heat equation with initial condition u(x, 0) = g(x) for -? < x < ? and 0 < t < ? by applying a convolution:

u ( x , t ) = ? ? ( x - y , t ) g ( y ) d y . {\displaystyle u(x,t)=\int \Phi (x-y,t)g(y)dy.}

In several spatial variables, the fundamental solution solves the analogous problem

{ u t ( x , t ) - k ? i = 1 n u x i x i ( x , t ) = 0 ( x , t ) ? R n × ( 0 , ? ) u ( x , 0 ) = ? ( x ) {\displaystyle {\begin{cases}u_{t}(\mathbf {x} ,t)-k\sum _{i=1}^{n}u_{x_{i}x_{i}}(\mathbf {x} ,t)=0&(\mathbf {x} ,t)\in \mathbf {R} ^{n}\times (0,\infty )\\u(\mathbf {x} ,0)=\delta (\mathbf {x} )\end{cases}}}

The n-variable fundamental solution is the product of the fundamental solutions in each variable; i.e.,

? ( x , t ) = ? ( x 1 , t ) ? ( x 2 , t ) ... ? ( x n , t ) = 1 ( 4 ? k t ) n exp ( - x ? x 4 k t ) . {\displaystyle \Phi (\mathbf {x} ,t)=\Phi (x_{1},t)\Phi (x_{2},t)\dots \Phi (x_{n},t)={\frac {1}{\sqrt {(4\pi kt)^{n}}}}\exp \left(-{\frac {\mathbf {x} \cdot \mathbf {x} }{4kt}}\right).}

The general solution of the heat equation on Rn is then obtained by a convolution, so that to solve the initial value problem with u(x, 0) = g(x), one has

u ( x , t ) = ? R n ? ( x - y , t ) g ( y ) d y . {\displaystyle u(\mathbf {x} ,t)=\int _{\mathbf {R} ^{n}}\Phi (\mathbf {x} -\mathbf {y} ,t)g(\mathbf {y} )d\mathbf {y} .}

The general problem on a domain ? in Rn is

{ u t ( x , t ) - k ? i = 1 n u x i x i ( x , t ) = 0 ( x , t ) ? ? × ( 0 , ? ) u ( x , 0 ) = g ( x ) x ? ? {\displaystyle {\begin{cases}u_{t}(\mathbf {x} ,t)-k\sum _{i=1}^{n}u_{x_{i}x_{i}}(\mathbf {x} ,t)=0&(\mathbf {x} ,t)\in \Omega \times (0,\infty )\\u(\mathbf {x} ,0)=g(\mathbf {x} )&\mathbf {x} \in \Omega \end{cases}}}

with either Dirichlet or Neumann boundary data. A Green's function always exists, but unless the domain ? can be readily decomposed into one-variable problems (see below), it may not be possible to write it down explicitly. Other methods for obtaining Green's functions include the method of images, separation of variables, and Laplace transforms (Cole, 2011).

Some Green's function solutions in 1D

A variety of elementary Green's function solutions in one-dimension are recorded here; many others are available elsewhere. In some of these, the spatial domain is (-?,?). In others, it is the semi-infinite interval (0,?) with either Neumann or Dirichlet boundary conditions. One further variation is that some of these solve the inhomogeneous equation

u t = k u x x + f . {\displaystyle u_{t}=ku_{xx}+f.}

where f is some given function of x and t.

Homogeneous heat equation

Initial value problem on (-?,?)
{ u t = k u x x ( x , t ) ? R × ( 0 , ? ) u ( x , 0 ) = g ( x ) I C {\displaystyle {\begin{cases}u_{t}=ku_{xx}&(x,t)\in \mathbf {R} \times (0,\infty )\\u(x,0)=g(x)&IC\end{cases}}}
u ( x , t ) = 1 4 ? k t ? - ? ? exp ( - ( x - y ) 2 4 k t ) g ( y ) d y {\displaystyle u(x,t)={\frac {1}{\sqrt {4\pi kt}}}\int _{-\infty }^{\infty }\exp \left(-{\frac {(x-y)^{2}}{4kt}}\right)g(y)\,dy}

Comment. This solution is the convolution with respect to the variable x of the fundamental solution

? ( x , t ) := 1 4 ? k t exp ( - x 2 4 k t ) , {\displaystyle \Phi (x,t):={\frac {1}{\sqrt {4\pi kt}}}\exp \left(-{\frac {x^{2}}{4kt}}\right),}

and the function g(x). (The Green's function number of the fundamental solution is X00.) Therefore, according to the general properties of the convolution with respect to differentiation, u = g * ? is a solution of the same heat equation, for

( ? t - k ? x 2 ) ( ? * g ) = [ ( ? t - k ? x 2 ) ? ] * g = 0. {\displaystyle \left(\partial _{t}-k\partial _{x}^{2}\right)(\Phi *g)=\left[\left(\partial _{t}-k\partial _{x}^{2}\right)\Phi \right]*g=0.}

Moreover,

? ( x , t ) = 1 t ? ( x t ) {\displaystyle \Phi (x,t)={\frac {1}{\sqrt {t}}}\,\Phi \left({\frac {x}{\sqrt {t}}}\right)}
? - ? ? ? ( x , t ) d x = 1 , {\displaystyle \int _{-\infty }^{\infty }\Phi (x,t)\,dx=1,}

so that, by general facts about approximation to the identity, ?(?, t) * g -> g as t -> 0 in various senses, according to the specific g. For instance, if g is assumed bounded and continuous on R then ?(?, t) * g converges uniformly to g as t -> 0, meaning that u(x, t) is continuous on R × [0, ?) with u(x, 0) = g(x).

Initial value problem on (0,?) with homogeneous Dirichlet boundary conditions
{ u t = k u x x ( x , t ) ? [ 0 , ? ) × ( 0 , ? ) u ( x , 0 ) = g ( x ) I C u ( 0 , t ) = 0 B C {\displaystyle {\begin{cases}u_{t}=ku_{xx}&(x,t)\in [0,\infty )\times (0,\infty )\\u(x,0)=g(x)&IC\\u(0,t)=0&BC\end{cases}}}
u ( x , t ) = 1 4 ? k t ? 0 ? [ exp ( - ( x - y ) 2 4 k t ) - exp ( - ( x + y ) 2 4 k t ) ] g ( y ) d y {\displaystyle u(x,t)={\frac {1}{\sqrt {4\pi kt}}}\int _{0}^{\infty }\left[\exp \left(-{\frac {(x-y)^{2}}{4kt}}\right)-\exp \left(-{\frac {(x+y)^{2}}{4kt}}\right)\right]g(y)\,dy}

Comment. This solution is obtained from the preceding formula as applied to the data g(x) suitably extended to R, so as to be an odd function, that is, letting g(-x) := -g(x) for all x. Correspondingly, the solution of the initial value problem on (-?,?) is an odd function with respect to the variable x for all values of t, and in particular it satisfies the homogeneous Dirichlet boundary conditions u(0, t) = 0. The Green's function number of this solution is X10.

Initial value problem on (0,?) with homogeneous Neumann boundary conditions
{ u t = k u x x ( x , t ) ? [ 0 , ? ) × ( 0 , ? ) u ( x , 0 ) = g ( x ) I C u x ( 0 , t ) = 0 B C {\displaystyle {\begin{cases}u_{t}=ku_{xx}&(x,t)\in [0,\infty )\times (0,\infty )\\u(x,0)=g(x)&IC\\u_{x}(0,t)=0&BC\end{cases}}}
u ( x , t ) = 1 4 ? k t ? 0 ? [ exp ( - ( x - y ) 2 4 k t ) + exp ( - ( x + y ) 2 4 k t ) ] g ( y ) d y {\displaystyle u(x,t)={\frac {1}{\sqrt {4\pi kt}}}\int _{0}^{\infty }\left[\exp \left(-{\frac {(x-y)^{2}}{4kt}}\right)+\exp \left(-{\frac {(x+y)^{2}}{4kt}}\right)\right]g(y)\,dy}

Comment. This solution is obtained from the first solution formula as applied to the data g(x) suitably extended to R so as to be an even function, that is, letting g(-x) := g(x) for all x. Correspondingly, the solution of the initial value problem on R is an even function with respect to the variable x for all values of t > 0, and in particular, being smooth, it satisfies the homogeneous Neumann boundary conditions ux(0, t) = 0. The Green's function number of this solution is X20.

Problem on (0,?) with homogeneous initial conditions and non-homogeneous Dirichlet boundary conditions
{ u t = k u x x ( x , t ) ? [ 0 , ? ) × ( 0 , ? ) u ( x , 0 ) = 0 I C u ( 0 , t ) = h ( t ) B C {\displaystyle {\begin{cases}u_{t}=ku_{xx}&(x,t)\in [0,\infty )\times (0,\infty )\\u(x,0)=0&IC\\u(0,t)=h(t)&BC\end{cases}}}
u ( x , t ) = ? 0 t x 4 ? k ( t - s ) 3 exp ( - x 2 4 k ( t - s ) ) h ( s ) d s , ? x > 0 {\displaystyle u(x,t)=\int _{0}^{t}{\frac {x}{\sqrt {4\pi k(t-s)^{3}}}}\exp \left(-{\frac {x^{2}}{4k(t-s)}}\right)h(s)\,ds,\qquad \forall x>0}

Comment. This solution is the convolution with respect to the variable t of

? ( x , t ) := - 2 k ? x ? ( x , t ) = x 4 ? k t 3 exp ( - x 2 4 k t ) {\displaystyle \psi (x,t):=-2k\partial _{x}\Phi (x,t)={\frac {x}{\sqrt {4\pi kt^{3}}}}\exp \left(-{\frac {x^{2}}{4kt}}\right)}

and the function h(t). Since ?(x, t) is the fundamental solution of

? t - k ? x 2 , {\displaystyle \partial _{t}-k\partial _{x}^{2},}

the function ?(x, t) is also a solution of the same heat equation, and so is u := ? * h, thanks to general properties of the convolution with respect to differentiation. Moreover,

? ( x , t ) = 1 x 2 ? ( 1 , t x 2 ) {\displaystyle \psi (x,t)={\frac {1}{x^{2}}}\,\psi \left(1,{\frac {t}{x^{2}}}\right)}
? 0 ? ? ( x , t ) d t = 1 , {\displaystyle \int _{0}^{\infty }\psi (x,t)\,dt=1,}

so that, by general facts about approximation to the identity, ?(x, ?) * h -> h as x -> 0 in various senses, according to the specific h. For instance, if h is assumed continuous on R with support in [0, ?) then ?(x, ?) * h converges uniformly on compacta to h as x -> 0, meaning that u(x, t) is continuous on [0, ?) × [0, ?) with u(0, t) = h(t).

Inhomogeneous heat equation

Problem on (-?,?) homogeneous initial conditions
{ u t = k u x x + f ( x , t ) ( x , t ) ? R × ( 0 , ? ) u ( x , 0 ) = 0 I C {\displaystyle {\begin{cases}u_{t}=ku_{xx}+f(x,t)&(x,t)\in \mathbf {R} \times (0,\infty )\\u(x,0)=0&IC\end{cases}}}
u ( x , t ) = ? 0 t ? - ? ? 1 4 ? k ( t - s ) exp ( - ( x - y ) 2 4 k ( t - s ) ) f ( y , s ) d y d s {\displaystyle u(x,t)=\int _{0}^{t}\int _{-\infty }^{\infty }{\frac {1}{\sqrt {4\pi k(t-s)}}}\exp \left(-{\frac {(x-y)^{2}}{4k(t-s)}}\right)f(y,s)\,dy\,ds}

Comment. This solution is the convolution in R2, that is with respect to both the variables x and t, of the fundamental solution

? ( x , t ) := 1 4 ? k t exp ( - x 2 4 k t ) {\displaystyle \Phi (x,t):={\frac {1}{\sqrt {4\pi kt}}}\exp \left(-{\frac {x^{2}}{4kt}}\right)}

and the function f(x, t), both meant as defined on the whole R2 and identically 0 for all t -> 0. One verifies that

( ? t - k ? x 2 ) ( ? * f ) = f , {\displaystyle \left(\partial _{t}-k\partial _{x}^{2}\right)(\Phi *f)=f,}

which expressed in the language of distributions becomes

( ? t - k ? x 2 ) ? = ? , {\displaystyle \left(\partial _{t}-k\partial _{x}^{2}\right)\Phi =\delta ,}

where the distribution ? is the Dirac's delta function, that is the evaluation at 0.

Problem on (0,?) with homogeneous Dirichlet boundary conditions and initial conditions
{ u t = k u x x + f ( x , t ) ( x , t ) ? [ 0 , ? ) × ( 0 , ? ) u ( x , 0 ) = 0 I C u ( 0 , t ) = 0 B C {\displaystyle {\begin{cases}u_{t}=ku_{xx}+f(x,t)&(x,t)\in [0,\infty )\times (0,\infty )\\u(x,0)=0&IC\\u(0,t)=0&BC\end{cases}}}
u ( x , t ) = ? 0 t ? 0 ? 1 4 ? k ( t - s ) ( exp ( - ( x - y ) 2 4 k ( t - s ) ) - exp ( - ( x + y ) 2 4 k ( t - s ) ) ) f ( y , s ) d y d s {\displaystyle u(x,t)=\int _{0}^{t}\int _{0}^{\infty }{\frac {1}{\sqrt {4\pi k(t-s)}}}\left(\exp \left(-{\frac {(x-y)^{2}}{4k(t-s)}}\right)-\exp \left(-{\frac {(x+y)^{2}}{4k(t-s)}}\right)\right)f(y,s)\,dy\,ds}

Comment. This solution is obtained from the preceding formula as applied to the data f(x, t) suitably extended to R × [0,?), so as to be an odd function of the variable x, that is, letting f(-x, t) := -f(x, t) for all x and t. Correspondingly, the solution of the inhomogeneous problem on (-?,?) is an odd function with respect to the variable x for all values of t, and in particular it satisfies the homogeneous Dirichlet boundary conditions u(0, t) = 0.

Problem on (0,?) with homogeneous Neumann boundary conditions and initial conditions
{ u t = k u x x + f ( x , t ) ( x , t ) ? [ 0 , ? ) × ( 0 , ? ) u ( x , 0 ) = 0 I C u x ( 0 , t ) = 0 B C {\displaystyle {\begin{cases}u_{t}=ku_{xx}+f(x,t)&(x,t)\in [0,\infty )\times (0,\infty )\\u(x,0)=0&IC\\u_{x}(0,t)=0&BC\end{cases}}}
u ( x , t ) = ? 0 t ? 0 ? 1 4 ? k ( t - s ) ( exp ( - ( x - y ) 2 4 k ( t - s ) ) + exp ( - ( x + y ) 2 4 k ( t - s ) ) ) f ( y , s ) d y d s {\displaystyle u(x,t)=\int _{0}^{t}\int _{0}^{\infty }{\frac {1}{\sqrt {4\pi k(t-s)}}}\left(\exp \left(-{\frac {(x-y)^{2}}{4k(t-s)}}\right)+\exp \left(-{\frac {(x+y)^{2}}{4k(t-s)}}\right)\right)f(y,s)\,dy\,ds}

Comment. This solution is obtained from the first formula as applied to the data f(x, t) suitably extended to R × [0,?), so as to be an even function of the variable x, that is, letting f(-x, t) := f(x, t) for all x and t. Correspondingly, the solution of the inhomogeneous problem on (-?,?) is an even function with respect to the variable x for all values of t, and in particular, being a smooth function, it satisfies the homogeneous Neumann boundary conditions ux(0, t) = 0.

Examples

Since the heat equation is linear, solutions of other combinations of boundary conditions, inhomogeneous term, and initial conditions can be found by taking an appropriate linear combination of the above Green's function solutions.

For example, to solve

{ u t = k u x x + f ( x , t ) ? R × ( 0 , ? ) u ( x , 0 ) = g ( x ) I C {\displaystyle {\begin{cases}u_{t}=ku_{xx}+f&(x,t)\in \mathbf {R} \times (0,\infty )\\u(x,0)=g(x)&IC\end{cases}}}

let u = w + v where w and v solve the problems

{ v t = k v x x + f , w t = k w x x ( x , t ) ? R × ( 0 , ? ) v ( x , 0 ) = 0 , w ( x , 0 ) = g ( x ) I C {\displaystyle {\begin{cases}v_{t}=kv_{xx}+f,\,w_{t}=kw_{xx}\,&(x,t)\in \mathbf {R} \times (0,\infty )\\v(x,0)=0,\,w(x,0)=g(x)\,&IC\end{cases}}}

Similarly, to solve

{ u t = k u x x + f ( x , t ) ? [ 0 , ? ) × ( 0 , ? ) u ( x , 0 ) = g ( x ) I C u ( 0 , t ) = h ( t ) B C {\displaystyle {\begin{cases}u_{t}=ku_{xx}+f&(x,t)\in [0,\infty )\times (0,\infty )\\u(x,0)=g(x)&IC\\u(0,t)=h(t)&BC\end{cases}}}

let u = w + v + r where w, v, and r solve the problems

{ v t = k v x x + f , w t = k w x x , r t = k r x x ( x , t ) ? [ 0 , ? ) × ( 0 , ? ) v ( x , 0 ) = 0 , w ( x , 0 ) = g ( x ) , r ( x , 0 ) = 0 I C v ( 0 , t ) = 0 , w ( 0 , t ) = 0 , r ( 0 , t ) = h ( t ) B C {\displaystyle {\begin{cases}v_{t}=kv_{xx}+f,\,w_{t}=kw_{xx},\,r_{t}=kr_{xx}&(x,t)\in [0,\infty )\times (0,\infty )\\v(x,0)=0,\;w(x,0)=g(x),\;r(x,0)=0&IC\\v(0,t)=0,\;w(0,t)=0,\;r(0,t)=h(t)&BC\end{cases}}}

New Page 1
src: www.math.ualberta.ca


Mean-value property for the heat equation

Solutions of the heat equations

( ? t - ? ) u = 0 {\displaystyle (\partial _{t}-\Delta )u=0}

satisfy a mean-value property analogous to the mean-value properties of harmonic functions, solutions of

? u = 0 {\displaystyle \Delta u=0} ,

though a bit more complicated. Precisely, if u solves

( ? t - ? ) u = 0 {\displaystyle (\partial _{t}-\Delta )u=0}

and

( x , t ) + E ? ? d o m ( u ) {\displaystyle (x,t)+E_{\lambda }\subset \mathrm {dom} (u)}

then

u ( x , t ) = ? 4 ? E ? u ( x - y , t - s ) | y | 2 s 2 d s d y , {\displaystyle u(x,t)={\frac {\lambda }{4}}\int _{E_{\lambda }}u(x-y,t-s){\frac {|y|^{2}}{s^{2}}}ds\,dy,}

where E? is a "heat-ball", that is a super-level set of the fundamental solution of the heat equation:

E ? := { ( y , s ) : ? ( y , s ) > ? } , {\displaystyle E_{\lambda }:=\{(y,s)\,:\,\Phi (y,s)>\lambda \},}
? ( x , t ) := ( 4 t ? ) - n 2 exp ( - | x | 2 4 t ) . {\displaystyle \Phi (x,t):=(4t\pi )^{-{\frac {n}{2}}}\exp \left(-{\frac {|x|^{2}}{4t}}\right).}

Notice that

d i a m ( E ? ) = o ( 1 ) {\displaystyle \mathrm {diam} (E_{\lambda })=o(1)}

as ? -> ? so the above formula holds for any (x, t) in the (open) set dom(u) for ? large enough. Conversely, any function u satisfying the above mean-value property on an open domain of Rn × R is a solution of the heat equation. This can be shown by an argument similar to the analogous one for harmonic functions.


Cooling of a thin metal plate (2D transient heat equation) - YouTube
src: i.ytimg.com


Steady-state heat equation

The steady-state heat equation is by definition not dependent on time. In other words, it is assumed conditions exist such that:

? u ? t = 0 {\displaystyle {\frac {\partial u}{\partial t}}=0}

This condition depends on the time constant and the amount of time passed since boundary conditions have been imposed. Thus, the condition is fulfilled in situations in which the time equilibrium constant is fast enough that the more complex time-dependent heat equation can be approximated by the steady-state case. Equivalently, the steady-state condition exists for all cases in which enough time has passed that the thermal field u no longer evolves in time.

In the steady-state case, a spatial thermal gradient may (or may not) exist, but if it does, it does not change in time. This equation therefore describes the end result in all thermal problems in which a source is switched on (for example, an engine started in an automobile), and enough time has passed for all permanent temperature gradients to establish themselves in space, after which these spatial gradients no longer change in time (as again, with an automobile in which the engine has been running for long enough). The other (trivial) solution is for all spatial temperature gradients to disappear as well, in which case the temperature become uniform in space, as well.

The equation is much simpler and can help to understand better the physics of the materials without focusing on the dynamic of the heat transport process. It is widely used for simple engineering problems assuming there is equilibrium of the temperature fields and heat transport, with time.

Steady-state condition:

? u ? t = 0 {\displaystyle {\frac {\partial u}{\partial t}}=0}

The steady-state heat equation for a volume that contains a heat source (the inhomogeneous case), is the Poisson's equation:

- k ? 2 u = q {\displaystyle -k\nabla ^{2}u=q}

where u is the temperature, k is the thermal conductivity and q the heat-flux density of the source.

In electrostatics, this is equivalent to the case where the space under consideration contains an electrical charge.

The steady-state heat equation without a heat source within the volume (the homogeneous case) is the equation in electrostatics for a volume of free space that does not contain a charge. It is described by Laplace's equation:

? 2 u = 0 {\displaystyle \nabla ^{2}u=0}

Inverse Heat Conduction Problem, Heatlab
src: www.heatlab.cz


Applications

Particle diffusion

One can model particle diffusion by an equation involving either:

  • the volumetric concentration of particles, denoted c, in the case of collective diffusion of a large number of particles, or
  • the probability density function associated with the position of a single particle, denoted P.

In either case, one uses the heat equation

c t = D ? c , {\displaystyle c_{t}=D\Delta c,\quad }

or

P t = D ? P . {\displaystyle P_{t}=D\Delta P.\quad }

Both c and P are functions of position and time. D is the diffusion coefficient that controls the speed of the diffusive process, and is typically expressed in meters squared over second. If the diffusion coefficient D is not constant, but depends on the concentration c (or P in the second case), then one gets the nonlinear diffusion equation.

Brownian motion

Let the stochastic process X {\displaystyle X} be the solution of the stochastic differential equation

{ d X t = 2 k d B t X 0 = 0 {\displaystyle \left\{{\begin{array}{ll}{\text{d}}X_{t}={\sqrt {2k}}{\text{d}}B_{t}\\X_{0}=0\end{array}}\right.}

where B {\displaystyle B} is the Wiener process (standard Brownian motion). Then the probability density function of X {\displaystyle X} is given at any time t {\displaystyle t} by

1 4 ? k t exp ( - x 2 4 k t ) {\displaystyle {\frac {1}{\sqrt {4\pi kt}}}\exp \left(-{\frac {x^{2}}{4kt}}\right)}

which is the solution of the initial value problem

{ u t ( x , t ) - k u x x ( x , t ) = 0 , ( x , t ) ? R × ( 0 , + ? ) u ( x , 0 ) = ? ( x ) {\displaystyle \left\{{\begin{array}{ll}u_{t}(x,t)-ku_{xx}(x,t)=0,\quad (x,t)\in \mathbb {R} \times (0,+\infty )\\u(x,0)=\delta (x)\end{array}}\right.}

where ? {\displaystyle \delta } is the Dirac delta function.

Schrödinger equation for a free particle

With a simple division, the Schrödinger equation for a single particle of mass m in the absence of any applied force field can be rewritten in the following way:

? t = i ? 2 m ? ? {\displaystyle \psi _{t}={\frac {i\hbar }{2m}}\Delta \psi } ,

where i is the imaginary unit, ? is the reduced Planck's constant, and ? is the wave function of the particle.

This equation is formally similar to the particle diffusion equation, which one obtains through the following transformation:

c ( R , t ) -> ? ( R , t ) D -> i ? 2 m {\displaystyle {\begin{aligned}c({\mathbf {R}},t)&\to \psi ({\mathbf {R}},t)\\D&\to {\frac {i\hbar }{2m}}\end{aligned}}}

Applying this transformation to the expressions of the Green functions determined in the case of particle diffusion yields the Green functions of the Schrödinger equation, which in turn can be used to obtain the wave function at any time through an integral on the wave function at t = 0:

? ( R , t ) = ? ? ( R 0 , t = 0 ) G ( R - R 0 , t ) d R x 0 d R y 0 d R z 0 , {\displaystyle \psi ({\mathbf {R}},t)=\int \psi ({\mathbf {R}}^{0},t=0)G({\mathbf {R}}-{\mathbf {R}}^{0},t)dR_{x}^{0}\,dR_{y}^{0}\,dR_{z}^{0},}

with

G ( R , t ) = ( m 2 ? i ? t ) 3 / 2 e - R 2 m 2 i ? t . {\displaystyle G({\mathbf {R}},t)={\bigg (}{\frac {m}{2\pi i\hbar t}}{\bigg )}^{3/2}e^{-{\frac {{\mathbf {R}}^{2}m}{2i\hbar t}}}.}

Remark: this analogy between quantum mechanics and diffusion is a purely formal one. Physically, the evolution of the wave function satisfying Schrödinger's equation might have an origin other than diffusion.

Thermal diffusivity in polymers

A direct practical application of the heat equation, in conjunction with Fourier theory, in spherical coordinates, is the prediction of thermal transfer profiles and the measurement of the thermal diffusivity in polymers (Unsworth and Duarte). This dual theoretical-experimental method is applicable to rubber, various other polymeric materials of practical interest, and microfluids. These authors derived an expression for the temperature at the center of a sphere TC

T C - T S T 0 - T S = 2 ? n = 1 ? ( - 1 ) n + 1 exp ( - n 2 ? 2 ? t L 2 ) {\displaystyle {\frac {T_{C}-T_{S}}{T_{0}-T_{S}}}=2\sum _{n=1}^{\infty }(-1)^{n+1}\exp \left({-{\frac {n^{2}\pi ^{2}\alpha t}{L^{2}}}}\right)}

where T0 is the initial temperature of the sphere and TS the temperature at the surface of the sphere, of radius L. This equation has also found applications in protein energy transfer and thermal modeling in biophysics.

Further applications

The heat equation arises in the modeling of a number of phenomena and is often used in financial mathematics in the modeling of options. The famous Black-Scholes option pricing model's differential equation can be transformed into the heat equation allowing relatively easy solutions from a familiar body of mathematics. Many of the extensions to the simple option models do not have closed form solutions and thus must be solved numerically to obtain a modeled option price. The equation describing pressure diffusion in a porous medium is identical in form with the heat equation. Diffusion problems dealing with Dirichlet, Neumann and Robin boundary conditions have closed form analytic solutions (Thambynayagam 2011). The heat equation is also widely used in image analysis (Perona & Malik 1990) and in machine-learning as the driving theory behind scale-space or graph Laplacian methods. The heat equation can be efficiently solved numerically using the implicit Crank-Nicolson method of (Crank & Nicolson 1947). This method can be extended to many of the models with no closed form solution, see for instance (Wilmott, Howison & Dewynne 1995).

An abstract form of heat equation on manifolds provides a major approach to the Atiyah-Singer index theorem, and has led to much further work on heat equations in Riemannian geometry.




See also

  • Caloric polynomial
  • Curve-shortening flow
  • Diffusion equation
  • Relativistic heat conduction
  • Schrödinger equation



Notes




References

  • Cannon, John Rozier (1984), The One-Dimensional Heat Equation, Encyclopedia of Mathematics and Its Applications, 23 (1st ed.), Reading-Menlo Park-London-Don Mills-Sidney-Tokyo/ Cambridge-New York-New Rochelle-Melbourne-Sidney: Addison-Wesley Publishing Company/Cambridge University Press, pp. XXV+483, ISBN 978-0-521-30243-2, MR 0747979, Zbl 0567.35001 .
  • Crank, J.; Nicolson, P. (1947), "A Practical Method for Numerical Evaluation of Solutions of Partial Differential Equations of the Heat-Conduction Type", Proceedings of the Cambridge Philosophical Society, 43: 50-67, Bibcode:1947PCPS...43...50C, doi:10.1017/S0305004100023197 
  • Einstein, Albert (1905), "Über die von der molekularkinetischen Theorie der Wärme geforderte Bewegung von in ruhenden Flüssigkeiten suspendierten Teilchen", Annalen der Physik, 322 (8): 549-560, Bibcode:1905AnP...322..549E, doi:10.1002/andp.19053220806 
  • Evans, L.C. (1998), Partial Differential Equations, American Mathematical Society, ISBN 0-8218-0772-2 
  • Cole, K.D.; Beck, J.V.; Haji-Sheikh, A.; Litkouhi, B. (2011), Heat Conduction Using Green's Functions (2nd ed.), CRC Press, ISBN 978-1-43-981354-6 
  • John, Fritz (1991), Partial Differential Equations (4th ed.), Springer, ISBN 978-0-387-90609-6 
  • Wilmott, P.; Howison, S.; Dewynne, J. (1995), The Mathematics of Financial Derivatives:A Student Introduction, Cambridge University Press 
  • Carslaw, H. S.; Jaeger, J. C. (1959), Conduction of Heat in Solids (2nd ed.), Oxford University Press, ISBN 978-0-19-853368-9 
  • Thambynayagam, R. K. M. (2011), The Diffusion Handbook: Applied Solutions for Engineers, McGraw-Hill Professional, ISBN 978-0-07-175184-1 
  • Perona, P; Malik, J. (1990), "Scale-Space and Edge Detection Using Anisotropic Diffusion", IEEE Transactions on Pattern Analysis and Machine Intelligence, 12 (7): 629-639, doi:10.1109/34.56205 
  • Unsworth, J.; Duarte, F. J. (1979), "Heat diffusion in a solid sphere and Fourier Theory", Am. J. Phys., 47 (11): 891-893, Bibcode:1979AmJPh..47..981U, doi:10.1119/1.11601 



External links

  • Derivation of the heat equation
  • Linear heat equations: Particular solutions and boundary value problems - from EqWorld

Source of the article : Wikipedia

Comments
0 Comments